Examination Puzzle

Date: 02/14/2003 at 20:41:44
From: Brian
Subject: Math puzzle
Below is a puzzle I found. I figured out the answer, but only by
trial and error.
3 students are taking an examination of N subjects. Points are
awarded according to the positions in each subject they took. The
person who gets 1st will get the highest points, followed by 2nd and
lastly 3rd. (So let 1st = X points, 2nd = Y points, 3rd = Z points.)
Student B obtained X points in English. Student A obtained 22 points
total. Students B and C obtained 9 points total each. How many
subjects did they take in the examination, and what are the values of
X, Y, Z?
The answer is N = 5, X = 5, Y = 2, Z = 1. I was told that this is the
only solution with nonzero integers. I was wondering how you would go
about solving this in a more mathematical way.
The only thing I could think of doing was to try to solve the system
of equations:
aX+bY+cZ = 22
dX+eY+fZ = 9
gX+hY+iZ = 9
for X, Y, and Z in terms of a through i, and hoping it would be
apparent from there what to do next. This just got insanely
complicated and I gave up.

Date: 02/21/2003 at 23:51:39
From: Doctor Peterson
Subject: Re: Math puzzle
Hi, Brian.
My solution involves intelligent trial and error, so there aren't too
many things to try.
I have to assume that in each subject there were no ties or other
problems, so that one got X, one got Y, and one got Z. Then in each
subject the total score is X+Y+Z, and the total of all scores is
N(X+Y+Z).
Since this total is 22+9+9 = 40, N and X+Y+Z must be a pair of factors
of 40.
The smallest value for X+Y+Z would be 1+2+3, since they must be
different. So the greatest N can be is 40/6, and N<=6. The only
possible values for N (factors of 40 up to 6) are 1, 2, 4, and 5. If
there were only one subject, B and C could not get the same total, so
that is out. We are left with
N X+Y+Z
--- -----
2 20
4 10
5 8
Now we can look at the details:
[1] B got X, the largest, in one subject
[2] A's total is 22 (which can't be as great as NX)
[3] B and C both got 9
Suppose that N=2. Then the average of X, Y, and Z is 10; but then
X>10, which contradicts [1] and [3].
Suppose that N=4. For B to get 9, X can't be larger than 9-(1+1+1) =
6. For A to get 22, 4X must be greater than 22, so X is 6. Then Y+Z =
4, and we have Y = 3 and Z = 1. But if A got 3 6's, the fourth score
must have been 4. So this is out too.
So N = 5. For B to get 9, X can't be larger than 9-(1+1+1+1) = 5. For
A to get 22, 5X must be greater than 22, so X is 5. Then Y+Z = 3; and
since they are different, Y = 2 and Z = 1.
The scores therefore are
A: 5+5+5+5+2 = 22
B: 1+1+1+1+5 = 9
C: 2+2+2+2+1 = 9
It works!
I could probably refine this, but it's too late to do any more. Is
this logical enough for you?
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/